Sunday, March 18, 2012

"Proof" of Fermat's Last Theorem

One problem with the proliferation of "open access" journals is the decrease in quality. A good example is this "proof" of Fermat's Last Theorem by a guy who seems to specialize in rather eccentric papers. This paper was passed around to great laughter at the van der Poorten memorial conference in Australia. (The list of keywords alone is funny to a professional mathematician.)

This journal - the Journal of Mathematical and Computational Science - and its editorial board should be ashamed of publishing this junk.

175 comments:

I had a long correspondence with one of these folks. A fine retired gentleman who made one mistake after another in an attempt to use basic high-school math to prove FLT. I kept telling him that although I would never say never, the tools that he was using were simply of too feeble mathematical content to be likely to prove anything. He persisted till I gave up answering his e-mails.

Should I infer that you are opposed to the basic idea of an open-access journal, or are you simply pointing out that open-access can lead to a proliferation of bad ones? (I happen to believe that open-access journals are a great idea, but they place ever more responsibility for integrity on the editors.)

I edit a true open-access journal - that is, free to both authors and readers - so evidently no, I am not opposed to them.

There are a number of problems with many open access journals as they are currently set up. One is that most such journals have no provisions for waiving the author fees for those without a grant or those in 3rd world countries where the fee to publish might represent an average year's salary.

Thanks for the amusing link. (Surely he is credible, look at how diligent his research review was, why, he asks us to "see [1] or [3] or [4] or [5] or [6] or [7] or [8] or [9] or [10] or [11] or [12] or [13] or [15] or [18] or [19] or [20] or [21]"!)

However, I don't know why you focused on the openness of the journal's access. Closed journals are just as capable of publishing junk. Just look at Chaos, Solitons & Fractals.

Indeed, one might properly regard the Chaos, Solitons and Fractals issue as also being caused by a departure from sane journal-funding strategies.

If Elsevier sold journals individually, loads of people would want Advances in Mathematics (for example) and pretty much nobody would want Chaos, Solitons and Fractals, and it would not seem like such a big deal at all.

Apparently the author was charged $100 for publishing this paper. Forty papers have been published so far, and $4000 is a decent profit for putting up a web site. (It's clear from this paper that the level of copy-editing provided is minimal, if it's nonzero at all.)

Sadly, I recognize several names on the editorial board, probably people who have not closely looked into the standards of this journal:http://scik.org/index.php/jmcs/announcement/view/5

I didn't realize your journal was open access. I'd be interested in chatting with you about that at some point, particularly since I just took over as editor of a non open access journal in January. Do you have a web page up that explains the mechanics (policies, payments, copyediting, etc) for the journal?

The South Asian Journal of Mathematics seems to be full of weird papers, such as http://www.sajm.com.nu/sajm2012_2_2_11gueye.pdf. Aside from what must be translation issues (switching between "prime" and "first"), some things just seem inexplicable. For example, n^2 is used to denote squaring, but n^* is used for cubing.

Well, his address is ...@ccr.jussieu.fr. Replacing this by www.ccr.jussieu.fr brings us to web page of a department which looks like computing support division. In other words, he's probably a computing officer (centre du calcul, in french, probably means computing [support]). So, my guess is, he's a computer hacker with lots of free time and/or good copy/paste skills.

If you find Andrew Wiles proof is difficult, Pl. read on the intenet(open access pdf)CMNSEM,Vol.2 ,No.3,March 2011(General proof),CMNSEM,Vol.1,No.6,September2010(correct and simple use of method of infinite desscent of Fermat, in Case.N=3)CMNSEM,Vol.Vol.1,No.3,April 2010(except trivial typos,wonderful proofs seem OK)Extremely short simple proof for n=7 in CMNSEM,VOl.1,No.9,December 2011,However, a mistake may be seen in n=5 case published in the same journal.

Thanks a lot for valuable information on open acesss journals and Fermat's last theorem.The Author,Ikrong Anouk Gilbert Nemrong has already proved Fermat's last theorem seperately-A Complete Simple Proof of the Fermat's last theorem,International Mathematical Porum,Vol.7,2012,np.20,953-971. The paer (wonderful one,if it is OK) in the Joiral of mathematical and computer science had been reviewed within ten days!

Ikorong Anouk Gilbert Nemron has already proved Dermat's last theorem seperately.(A Complete Simple Proof of Fermat's Last theorem,International mathematical Forum,Vol.7,2012,no.20,953-971),This is not Mathematics,I guess,but Logic,etc.. I don't understand.Another one.A Simple Proof and Short Proof of Fermat's last Theirem,Fayez Fok Al Adel,Advances in Applied Mathmatical Analysiss,Vol.3,No.1 (2008)7-15,I never seen any comment on this simple proof,Even though author asked to comment on this proof seperately.This year I did not see it on the internet,I don't know why. Journal is open or not , the correct proofs must be appreciated and recimmended by people like you since mathematically sound Andrew Wiles-Taylor proof is extremely difficult and lengthy. Fermat's marvelous proof(?) is also available noW. Author is a professor.Daniele De Pedis Anyway, your information is useful to Honest Researchers. Thank you.

It is, A proof of the Goldbach conjecture and the strong attachment to the Fermat's last conjecture by Ikrong....NEMRON ,the paper on the South Asian Journal of Mathematics, 2011,Vol.1,No.2,68-80 had been reviwed within ten daye. I am soory, it is not the journal of mathematical and computational science.Thanks to you , anyway, we know that all big problems in mathematics have solved!!.Thank you also for your cooment on the same person's proof of Fermat's last theorem.

An Anonymous (not me) has reported the notice about the pubblication of my demostration of Fermat's Last Theorem. If you are interested in, you can found it on http://arxiv.org/ftp/arxiv/papers/1105/1105.0669.pdf Any comment/suggestion are welcomedaniele.depedis@roma1.infn.it

Publishing of nonsense in the open access journals seem to be possible then and has been done as well.This acctually kill the moral of honest reserachers who do not want fame or anything.There are many valuable results in open access journals which we can study freely.This must be apprected while hackneying nonsense,

I'm sorry to come rather late to this discussion, but the "South Asian Journal of Mathematics" has just been drawn to my attention via another route, and a web search for that journal led here.

Over the last few years various vested interests have promoted the "author pays" model of publication under the weaselly phrase "open access" (or even "gold open access"). There have been warnings that the adoption of such a model would lead to a rash of poor papers being published in fly-by-night journals. What I have only now realised is that this is already happening.

I think no researcher will be interested in commenting on the so called elementary proofs of Fermat;s last theorem not given in refereed journals.It is utter waste of time.Many of these authors do not know Fermat riples do not satisfy Pythagoras' equation.This is, by no means a prof of Fermat's last theorem.Unfortunately, some of these proofs have been given by Professors.

Prof.Fayez Fok Al Adeh's proof of Fermat's last theorem is not simple at all to me,a university teacher, which has been published in the Journal of Advances in Applied Mathematucal Analysis. However,I am trying to understand it.

Your counterexample is not valid, of course the A, B and C variables must fulfill the 's equation! On the other hand, the property gcd(B+A, C-A)=1 is reported also in the P.Ribbemboin book- "Fermat's last theorem for amateur".Clearly the statement must be interpreted as: "if there were three integers that satisfy the Fermat's equation then ..."

In any case, recently I have posted on http://arxiv.org/pdf/1105.0669.pdf a new version of my proposed "proof" much more simpler and general, any comment and/or suggestion are welcome.

On 12:21 PM, August 13, 2012 I have posted my answer to Paolo Tassotti concerning my "proof" of the Last Theorem. In that post,due to my inexperience about this blog, appear two errors:1) The author of post is Master instead of my name "Daniele De Pedis"2)on the first row is missing the name, that is:"...must fulfill the 's equation!" instead of "...must fulfill the 's equation!"

By reading this Blog(very very yseful), I understand that CMNSEM(Canadian Journal of Computing in mathematics,Natural Science,Engineering and Msrdicine) does not contain the general proof of Fermat's last theorem. This is rather unfortunate.Simple and short analytical proof of Fermat's lasttheorem is there.This will certainly help to reduce the calculation of Prof.Daniele De Pedis(Wrong!) proof drastically.Would you please point out a simple proof of Fermat's last theorem for n=3 case that I( University firstyear student of a third world country) can understand.Thank you very much regarding your comment on CMNSEM as well.

Let me join this discussion since I'm also engaged in despicable business of proving FLT. Here is an address of the abstract of the offered proof. Just in case if it may interest somebodyhttp://www.fileden.com/files/2012/5/4/3300464/Abstract%20of%20short%20Proof%20PDF%281%29.pdf

I don't argue with the people. Howwever,proving Pythagorean triples do not satisdy the Fermat equation(General)eqyation is not the proof of Fermat's last theorem. No one can nullify the papers I have published in CNMSEM.

What I did?First, i heard and read that no simple prrof for Fermat's last theorem for n=3.At the same time ,i understood that correcting Euler's proof (n=3 case)is very difficult.Secondly ,my aim was to give a simple proof for the general case.Finally , I wanted to know how peiple look at Andrew Wiles Proof.Taking these into account I did work on tme theorem for about three years.I think the proofe given in CMNSEM are sufficient at present,I thank you very much for the following,(1) Anonymous can ask you queations(2) Your commnet on CMNSEM is fare while others 'Killing' it. (3) Above all you are a proffesor in Canada I thank you again.

Fermat's last theorem has been the most difficult and the most famous theorem in mathemtics.I have read the proofs of some people.All I read that unpublished in a refereed journal are incorrect.It is nonsense to argue with the authors of these proofs to my mind, since I had experienced once .Proof developed in CMNSEM is OK and based on simple mathematics.Any one can challenge against the validity of the proof.

I don't want to be too much annoying in debunking De Pedis's "proof", probably just a waste of time.However, he tries to construct a univariate polynomial from the original trivariate statement saying "for each choice of the two other variables".It could be legal unless, some paragraphs below, he mixes up the FLT's trivial solution with this polynomial having a single integer root.This is clearly false because of the "for each choice of the other two variables" statement.

In any case he never use the property of the exponent "p" of being prime, and doesn't split the problem in the classic two cases (p | abc of not).He neither takes into account the divisibility property of the binomial coefficient.All symptoms of an hopeless attempt.

I don't understand the objection on first part of your post. Please could you "waste a little of your time" to explain better?

About the second part, the answers at all your objections are reported (as said in footnote 2) in the Rimbenboin book.

If you have found any error in my proof posted on arXiv.org please report it to me in clear statments, also by private email, if you wish, I will be grateful a lot!"All symptoms of an hopeless attempt" are not proof of fault !!

I'm again asking to look at concise (1 page) description of 4 steps of offered proof of FLT. Any basic flaw will be seen from it immediately. Its URLhttp://www.fileden.com/files/2012/5/4/3300464/Abstract%20of%20short%20Proof%20PDF%281%29.pdf

Regarding the South Asian Journalof Mathematics, a number of mathematicians, including A. Razborov,A. Abebe and N. Billor werelisted as editors by the journalwithout their knowledge or consent. (Their names have now been removed).

The extreme breadth of the journal's coverage; the small editorial board with tiny geographical coverage and unimpressive reputations; the fact that reviews are done in two weeks; the use of Microsoft Word as the text preparation software; -- all these are red flags that something is rather odd about this journal.

I have published all of my papers on Fermat's last theorem in CMNSEM. I actually did not know any bad thing about CMNSEM.In n=5 (with one of my students) we have made a mistake.Apart from that any one can challenge the validity of the proofs.

We have corrected the case n=5 and now we have two proofs of Fermat's last theorem (FLT)for n=5 using the Identity of Fermat equation, x+y-z.A fox, i have no any suitable word to call this man, has sent an E-mail to the authorities of my University saying that CNMSEM is not legtmate. Would you please tell us the open access journal that you edit.We are looking for the second simple proof of FLT similar to n=7 case in CMNSEM with one of my colleagues.R.A.D.P

I must indorm everyone who is interested in Fermat's last theorem that,mainly, the Dean of the university of Kelaniya has decided that CNMSEM is not a refereed journal.I know that the proof of FLT in CNMSEM is correct.However, anyones comment is appreciated.

It is the Dean of the science faculty, professor in mathematics,(I dont know what he knows) as far as I know that wanted to nullify all the papers related to FLT publisshed in CMNSEM by a lecturer of the department of mathematics of the University of Kelaniya.This is nasty and disgusting practice of this man.Let us see what will happen to the journal and proofs of FLT for many exponents.

Fermat,s last theorem(FLT) is the most notorious,difficult and famous theorem.If someone make(in finding a simple proof) a mistake in proving it there are a lot to laugh at it.Do these people appreciate the good work of it?.No.This is the truth.

Harvey Friedman grand conjecture says simple mathemtics canprove Fermat's last theorem. Also ,Colin Mactarty (philosopher)has reached the same conclusion.One has proved Fermat's last theorem for all exponents using elementary mathematics and published in CMNSEM.Anyone can challenge the world against the validity of the proofs on behalf of the author of the papers.

To prove Fermat's last theorem one needs Fundamental theorem of Arithmetic and Remainder theorem (A special form) only.R.A.D.P.Structure of Fermat triples in Mcpgr proof and proof in CMNSEM Vol.1,No.2,March 2011,pp57-63 are ninety percent coincide.One set in step.1 is wrong in Mcpgr ,therefore.I am reading the full proof however.

By now,I have read four proofs of Fermat's last theorem.Unfortunately, all are wrong.I have given up reading such proofs for sometime."A simple and short analytical proof of Fermat;s last theorem 'is available now.CMNSEM,Vol.2,No3. March 2011,p.57-63.Using the fundamental theorem of Arithmetic the theorem can be proved justifying Harvey Friedman conjecture and Colin McLarty prediction.However, by no means this underestimate Andrew Wiles work,to my mind. I thank you very much for giving me a very good chance to tell the above to the world.R.A.D.P

Investigate one paper, and you find a network of publishers and authors working together, http://www.journalshub.com/all-journal.php, example: American Journal of Math .. http://www.journalshub.com/journal-detail.php?journals_id=127 then google for Diploma Mills, Vanity Press, and cross reference the references.

I predicted "the death of the university" after being failed on a CELTA course in 2008.

A few weeks ago, I noticed that the rentacoder site was offering work for an assignment to be written. I e-mailed the lecturer to say that I did not want to make trouble for the person using rentacoder, but that this signalled to me the decline of civilisation as we know it.

I am very puzzled by blogs I have seen today on Fermat's Last Theorem.

There is Orwellian mention of an "open access journal". Does this mean that the journal is free to access on internet? I see now evidence that such a journal would be of poor quality.

50 years ago, we were told "publish or perish". We knew that 90% of the papers being published were rubbish.

Perhaps this is a blessing, as there is less to read. So it may be even more of a blessing if it turns out that at 99% of what is published today is rubbish.

Reading a blog on Fermats's Last Theorem. I can't get from one line to the next of Wiles's paper.But these blogs have at least the advantage of being easy to read. Maybe Wiles could republish his work in a way that is accessible to someone with only high school maths.

Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).

By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X”^p + Y”^p ?= Z”^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X’’ ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.

X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:

i) In (8), (9), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1),that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY”^(p-1) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.Fermat’s last theorem is simply proved!

ii) With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible.Fermat’s last theorem is simply again proved, with the connection to the concept of(X”/Z”)^p ?= 2. Is it interesting?

I suddenly thought of Fermat's last theorem at a night without sleep, while studying Quantum Physics and Maxwell equations years ago. And I believe and am rather sure that Fermat's last theorem was not just a hard-to-solve problem like many of us had thought of.

In fact, Fermat's last theorem is a base for setting up calculation methods for Quantum Physics and Wave equations if we merge the concept of continuity limit of real irrational numbers into rational numbers.

Please just correlate the notion of integers, rational numbers (the quantum is one, or some rational number!)to the basic concept of Quantum Physics (quantum-quanta).

And the concept of wave tranmission to the concept of continuity of real-irrational numbers!

If we apply the reasonings in the above proof of Fermat's last theorem to real-irrational numbers, we of course shall find out that the current concept of real-irrational numbers is still to be re-examined!

That is interesting, I think, because the concept of real-irrational numbers is basic for number theory, geometry, differential equations,trigonometry,...

Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).

By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X”^p + Y”^p ?= Z”^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X’’ ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.

X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:

i) In (8), (9), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1),that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY”^(p-1) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.Fermat’s last theorem is simply proved!

ii) With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible.Fermat’s last theorem is simply again proved, with the connection to the concept of(X”/Z”)^p ?= 2.

Sinh,In your "proof", nowhere do you need to use the fact that X,Y and Z are integers. Therefore, your proof also shows that X^p + Y^p = Z^p has no solutions when p>2 is prime and X, Y and Z are arbitrary real numbers! This is certainly false, therefore, your "proof" must be incorrect.

It is very interesting to discuss this. I wrote to Jeffrey Shallit that we can also apply this proof to real-irrational numbers, and find out abnormal new results for real-irrational number.

When the proof can be applied to whatever type of number, it is correct with that type of number! The main things to be checked are in the logical reasonings and the variables, too.

Please read:

ii) With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible.

With your question, we can rewrite that:

With X" and Z" are rational numbers (they come from the definition that X,Y,Z are integers), the equal sign in (X”/Z”)^p ?= 2 is impossible.

(The problem (X”/Z”)^p ?= 2 has been one of the basic excercises to be proved in number theory and to be examined for the concept of real-irrational numbers).

I am happy and delighted that I can disscuss Fermat's last theorem with you.

WE WOULD HAVE A LOT TO DISCUSS IN NUMBER THEORY, GEOMETRY, DIFFERENTIAL AND INTEGRAL CALCULUS,..., DIFFERENTIAL EQUATIONS,..., ALGEBRA,...,TRIGONOMETRY,..., QUANTUM PHYSICS,...ETCI CAN ASSURE YOU THAT!

(And I still avoid working with complex-imaginery numbers. I have different thoughts in definitions of complex-imaginery numbers!)

If you are still in doubt that the current concepts of real irrational numbers are still to be re-examined, let me post an exercise of real numbers that I have read in a number textbook as follows:

Let a,b,c,d are rational numbers, e is real irrational number, prove that if a+ce= b+de, then we must have a=b and c=d.

Proof:(a-b) ?= (d-c)e. Then we can easily see that a=b and c=d, otherwise e ?= (a-b)/(d-c). An irrational number cannot be equal to a rational number by definition.

We can see that, even with basic calculations such as addition and multiplication, irrational numbers behave not quite the same as rational numbers . This exercise reminds us the independence concept of vectors in linear algebra.

Then, the posted proof of Fermat's last theorem shows us that there is actually a new hidden difference between rational and irrational numbers.

If we want to deny a proof, we must point out the false and illogical reasoning or false, illogical variables used in the proof.We cannot arbitrarily apply a proof or function used for rational numbers to irrational numbers, right?

I wonder if there is proof that FLT can't be proven using goniometry. Otherwise that seems to me a much simpler way of approaching this problem.Given any value of a, b and n>2, the angle between side a and b is always somewhere between 60 and 90 degrees, so at least this gives you an illustration of the problem, while you can also do some math with this using the cosine law.

Why may I ask are mathematicians so adamant that there is no relatively simple proof of FLT just because it does not yield easily to their abstract mathematics when the maths of Euclid is all that is needed.We read that Pierre de Fermat had a 'EUREKA' moment and generated the equation z^n = y^n + x^n which now bears his name. NOT TRUE!! This was a creation by subsequent mathematicians as confirmed by the following statement of Andre Weil,: “Only on one ill-fated occasion did Fermat ever mention a curve of higher genus x^n + y^n = z^n ....... ".Fermat's revelation was that the algorithm for splitting a power x^n was identical with that for splitting the square so that he ended up with the generic equation x^n = z^2 - y^2 which is valid for EVERY INTEGER FROM 1 TO INFINITY. It is as plain as the nose on ones face for any idiot to see that the only exponent that will make all 3 exponents the same is 2 so his statement and hence his theorem is proved. The above algorithm is none other than Euclid's Elements, book2, props 5&6, available to Fermat and the basis for my simpleton reconstruction of Fermat's proof which is posted on U-tube. Why I ask has prop 8 of the same book been demoted to a rule, 'the quarters square rule' by mathematicians and then surreptitiously resurrected as Euclid's numbers when it suits them to explain Pythagorean triples. In case the mathematicians don't know it prop 8 is just 5&6 with a twist and actually more useful when it comes to explaining the powers.What do the binomial polynomials centred around Pascal's triangle tell the mathematicians? Nothing? What do the factored polynomials alias the 'Binomial Numbers' tell the mathematicians? Again apparently nothing.What they do tell one is that the basic property of the integers is 'unity' which is 'oneness' and of the powers it is 'squareness ' which is 'fourness' and that the progression from one level of exponentiation to the next is smooth and precise. And NO, it didn't require the modern 1993 proof to lay to rest the myth that somewhere out there in number space was a super massive 'Fermat Equation' waiting to jump out at us and cause a 'shockwave' of a tsunami' in the mathematical community as no mathematician worth his salt ever believed there was.

May I take this opportunity to thank you for posting my previous comment.Now let me think. The smallest pythagorean triple is 3^2 + 4^2 = 5^2 but hang on 4 is 2^2 so 4^2 is just (2^2)^2 which is umm.. 2^4 I think. Yes that's right. Now 2^4 is 2(2^3) which I believe is a rectangle of size 2 units by 8 units = 16 units which must be right since 4^2 equals 16. Now if I take 4 rectangles and form a square from them the square will be.. let's see... 8 +2 =10 ... that's right it will be 10^2 = 100 square units. Now the resulting internal square will be .... ehh .. 8 - 2 = 6 .... will be 6^2 = 36 square units. So 10^2 - 6^2 = 100 - 36 = 64. Now press the square root key on my Casio calculator and ... hey the answer's 8 which is a perfect square so 10^2 - 6^2 = 8^2.Now doesn't the 'Difference of Squares' formula a^2 - b^2 =(a+b)(a-b) [EE,book2,props5&6] so that 10^2 - 6^2 = (10+6)(10-6) =(16)(4) = 64 =8^2 where the quantities in brackets look remarkably like the 4 rectangles I started with!Now isn't forming 4 rectangles into a square just the 'Quarter Squares Rule' ... sorry I mean EE,book2,prop8 so that (8+2)^2 - (8-2)^2 = (4)(8)(2) =64 which of course is again just 10^2 - 6^2 = 8^2 which is just (4)5^2 - (4)3^2 = (4)4^2. The 'Quarter Squares Rule' should properly be called the 'Four Squares Rule' don't you think? If however we evaluate the 'QSR' as ((8+2)/2)^2 - ((8-2)/2)^2 we get 5^2 - 3^2 which as we all know equals 4^2 so I am led to believe that EE,book2,props5&6 and EE,book2,prop8 are just two sides of the same coin. Nay, not believe, THEY ARE one and the same thing.So 4^2 = 5^2 - 3^2 which is just the smallest pythagorean triple rearranged is identical with 2^4 = 5^2 - 3^2 which if my algebra is up to the task is just ... x^n = z^2 - y^2 ... and if my understanding of algebra is correct must apply to every integer from 1 to infinity. Hey, if I write x^n as (x^(n/2))^2 then I get ... (x^(n/2)^2 = z^2 - y^2 ... but isn't this just PYTHAGORUS'S EQUATION. So how can I elevate ... x^n = z^2 - y^2 ... into ... x^n = z^n - y^n ... I know I'll ask an expert, a mathmagician ... " You can't Jimmy" ... "Why?" ... " Because the 'Modern Proof of 1993' say's so" ... "Funny I thought it was Fermat circa 1630 that said so" .. " Oh no they're adamant that he didn't have the least idea what he was talking about".Well since the mathmagicians don't want to acknowledge the validity of ... x^n = z^2 - y^2 ... then I herewith claim priority for the algebraic form and hereby name it 'Bateman's Generic Algebraic Equation for the Triples of the Powers'. NOT BAD FOR A CRANK WOULDN'T YOU SAY!!Q.E.D. Proof by demonstration.

We beg to differ but as far as I am concerned Fermat said ”It is impossible to separate a cube into two cubes ......" which for me means that ... x^n = z^n - y^n ... because he found out that ... x^n = z^2 - y^2 ... prohibited it!! ' C'est la vie' and thank you for the comments.

Sorry for the typo but my last comment should have said ....We beg to differ but as far as I am concerned Fermat said ”It is impossible to separate a cube into two cubes ......" which for me means that ... x^n ≠ z^n - y^n ... because he found out that ... x^n = z^2 - y^2 ... prohibited it!! ' C'est la vie' and thank you for the comments.

It can be proved that a=nuwv+nw^n; b=nuwv+v^n; c=nuwv+w^n+v^n to satisfy equation a^n+b^n=c^n. When n=2 it becomes an identity (2wv+2w^2)^2+(2wv+v^2)^2=(2wv=2w^2+v^2)^2 (with u=1). For powers >2 it becomes a polynomial equation that maybe can proved having no integral roots by experienced mathematician. I proved it maybe not the best way.

It can be shown that to be integrals in a^n+b^n=c^n it's required a=nuwv+nw^n; b=nuwv+v^n; c=nuwv+nw^n+v^n. When n=2 we obtain identity (2wv+2w^2)^2+(2wv+v^2)^2==(2wv+2w^2+v^2)^2 (with u=1). When n>2 it becomes a polynomial equation that I believe can be proved not having integral roots by proficient mathematician . I maybe proved it not in the most elegant way

I wonder why when you are just in doubts of some simple proofs, you can say that such proofs are or will be proven wrong. We, of course, are often in doubts of many things, not only mathematics, and such doubts are necessary!

Moreover, if a technique is puny, it is right or wrong, not by being puny or not being puny. My attempted simple proof of Fermat's last theorem has been presented to be analyzed, and I should be very glad if you give me more analyses to the proof by pointing the wrong reasonings in the theorem.

Otherwise, we will always make a haste remark to any proof of any theorem that such proof is proven wrong just by saying that the proof is a)puny and b) doubtful.

Such haste remarks and conclusions are, of course, sometimes right and sometimes wrong, but they are not the way mathematicians apply in their work! is it essential in mathematics?

I wonder, why when you are in doubt of some proof, you think that such proof may be wrongly proven by just saying that it is a) puny and b) doubtful.We, of course, are often in doubts of many things in life and it is necessary.

My attempted simple proof have been presented to be analyzed and I should be very glad if you would give it more analyses.

Please give me more review to the reasonings of the proof.1+2 ?=3 is puny, but is it a essential part of mathematics?Thank you very much.

No, thanks. I have a limited amount of time to spend on things for which the prospect of success is so tiny. There are many, many amateurs who think they have proved Fermat's last theorem. So far, every amateur has failed. With such puny methods, the chances you have succeeded are virtually nil.

I would have loved a reply to my PS simply because there is a simple numerical recursive algorithm for the differences of adjacent numbers raised to the same power starting at a^0 which if fed back into the algorithm gives the differences between adjacent numbers raised to the next higher power and so I was wondering whether this would have any implications for FLT for the mathematicians. So the challenge for anyone wishing to take it on is to find the algorithm starting with nothing more than a column of zero's. By applying some simple algebra to the results it is possible to derive the polynomials for the 'Binomial Numbers'... see 'Wolfram MathWorld'... which could fit the bill for the polynomials that result from the identities given by McPgr. I will if no one succeeds supply the algorithm.

I guess there's only one way to silence the debate about simple proofs of FLT ones and for all: that is proving that for some principle reason it can't be done the easy way, let's say by proving that you can't make something like a distuingishing "cosine law for triplets" that could then be compared with FLT.

Perhaps we should be reminded that in the time that Fermat was living, the notion and concept of irrational numbers did not exist or were very primitive.

Mathematicians then built up irrational numbers and still now disregard the eery behaviour of irrational in the problem such as a+be ?= c+de (a,b,c,d: rational,e: irrational) and then treat calculus of irrational numbers in addition, subtraction, multiplication and division as the same as to calculus of rational numbers.

I know that almost all mathematicians (and very much possibly including me) have failed in the attempts of proving Fermat's last theorem. But I should be very glad when someone would point out the wrong reasonings in the proof. Otherwise, we would just again reject and deny any proof by just thinking that it is puny and doubtful.

Wiles proof is related with elliptic curves. We can see that the equation in my proof has a form:Y'^p ?= pX'(p-1) +...+1. Is it related with elliptic curves somehow?

Jeffrey Shallit said...."There are many, many amateurs who think they have proved Fermat's last theorem. So far, every amateur has failed. With such puny methods, the chances you have succeeded are virtually nil. ....."

That is all except mine of course since my proof is none other than PYTHAGORUS'S EQUATION and at the end of the day all the modern proof has to say is that Pythagorus's theorem reigns supreme but we mere mortals knew that all the time!!!

In response to Bert Brouwer's ongoing negative and senseless remarks I would suggest that he gets his behind out of his chair and prove his own banal offering but he should understand that his resulting drivel won't baffle, confuse or put anyone off their self established truths!!

Alastair, have you noticed that nowhere in your "proof" do you use the hypothesis, n > 2? That means you have succeeded in proving that there are no relatively prime integers such that a + b = c, a truly remarkable result.

Personally I feel perfectly happy to sit back and relax and watch this wild goose chase for many decades to come. But if Mr. Shallit himself wants to prove his own 'theorem' that some puny technique won't do the trick, because it doesn't distinguish adeqautely between the properties of integers and reals, then I would of course be glad to know.

Motherf***** of god... I'm sorry, am I not allowed to curse on your blog Dr Shallit. I blew a gasket reading whatshisname's elementary "proof" of FLT.

This idiot reminds me of an elderly Indian gentleman I know down the block who pretends he is a great wise physics-poet sage. The ironic thing is, he actually had earned some degree in physics, long ago, 50 years ago, enough to get a lifelong job in applying physics to the medical field: developing CAT scans & PET scans.

But, I can tell, he never had beyond calculus 50 years ago, of which he remembers nothing.

And his poetry sucks, too.

Anyway, I've gotten so jaded & now fearful of ALL peer-reviewed journals, fearing NONE of them seriously peer review works, that I would not know to which ones to submit a piece of math (which I don't have at present anyway).

In which journal did Wiles publish his proof of the Taniyama-Shimura Theorem? Surely THAT journal has to be reputable, no?

My question to amateurs who think they have proved all the big name theorems (by elementary means) or outstanding big name conjectures that are constantly mentioned throughout pop media, without FIRST going through the rigors of proving much LESS glamorous much LESS ambitious theorems along the way?

There are HUNDREDS of unsolved "lesser" conjectures in math, needing & deserving to be solved.

What makes you think that math does NOT require the long years of practice & trial & error & building up & learning from what others have done, doing the tedious boring exercises?

Why would you think math is the exception? Why would you not think that about playing sports or mastering a musical instrument or becoming a five-star general in the army after your first day in basic training?

To Bert Brouwer (I assume & hope no relationship to the famous Luitzen Brouwer who proved Brouwer's Fixed Point Theorem):guess who will be first to find an elementary proof of FLT?

Andrew Wiles: the one who first proved FLT the long slow hard highly advanced technical way. If ANYONE would know how to find an elementary proof of FLT, it will be Andrew Wiles & Gerhard Frey & Richard Taylor.

IThinkWithMy Liver can also sit back and relax, because I'm indeed not related to Luitzen Brouwer.I notice that he has already answered his own question he asked me:Guess who will be first to find an elementary proof of FLT?I will take his answer on faith, so I guess the only thing left for me to do is to say "Amen" to that... although (since it is a believe without proof) one can never know for sure of course. Who knows, maybe one day a martial arts instructor with a good spatial visualization ability and some elementary knowlegde of math is up to the task as well.

Thank you for your comment Gerry and for the slap on the back of the head which I needed since 'Reductio Ad Absurdum' is actually 'Proof Ad Absurdum' because the difference equations either side of the equality do not require the numbers to be the same i.e. 900 - 800 = 600 - 500 is valid. So my insisting that the difference of two squares should be the difference of the same two squares is rubbish and it is valid for it to be equal to the difference of two rectangles which have a common side. However the foregoing has no bearing on the validity of the equation x^n = z^2 - y^2 ... which is a proof of FLT as far as I am concerned.

Here as promised is the simple numerical recursive algorithm for the differences of the powers starting with nothing but a column of zeros which result from (z^0 -y^0) =(1-1) =0 which we then feed into the following algorithm. The resulting answers are feed back into the algorithm after the exponent of the first term has been increased by 1.For (z^1 -y^1/(z-y) : 0^0 +1x0 =1: 1^0 +2x0 =1: 2^0 +3x0 =1: 3^0 +4x0 =1: 4^0 +5x0 =1:For (z^2 -y^2)/(z-y) : 0^1 +1x1 =1: 1^1 +2x1 =3: 2^1 +3x1 =5: 3^1 +4x1 =7: 4^1 +5x1 =9:For (z^3 -y^3)/(z-y) : 0^2 +1x1 =1: 1^2 +2x3 =7: 2^2 +3x5 =19: 3^2 +4x7 =37: 4^2 +5x9 =61:For (z^4 -y^4)/(z-y) : 0^3 +1x1 =1: 1^3 +2x7 =15: 2^3 +3x19 =65: 3^3 +4x37 =175: 4^3 +5x61 =369:So what we have produced are the first differences of the finite difference tables and it is plain for ANYONE to see that these can be simply extended to apply to the infinity of powers applied to the infinity of integers. From the cubic onwards the differences are divergent which after feedback become more and more and more divergent. Now (z^n -y^n) = d1+d2+d3+... dm; that is the sum of the differences between z^n & y^n so why would anybody expect d1+d2+d3+... dm, the partial sum of a divergent series to equal x^n. Apparently the 'mathmagicians' did when they did their computer searches upto an exponent size of 4 million.Incidentally, the factored polynomial equations which are the binomial numbers can be derived directly from the arithmetic which exhibits the duality to be expected from interchange of the x's & y's i.e. 3^3 +4x37 =175 = 4^3 +3x37. I'll leave the algebra to you.Also from the 'DoS' equation we have the two identities as solutions to 'Fermat's' equation [1] {(a^2)^n -(b^2)^n}/(a^n-b^n) =(a^n+b^n) & [2] {(a^2)^n -(b^2)^n}/(a^n+b^n) =(a^n-b^n) neither of which look to be likely candidates for c^n do they?True , the foregoing is only factual observation and not analytic proof but it does give a lay person like myself an understandable insight into what is going on, which the modern proof doesn't and never could, plus the knowledge that contrary to the denial and attempted ridicule by the disciples of 'modern abstract mathematics', the tools and knowledge did exist for a person of Fermat's ability to prove his now famous statement.Also I ask, "Why did it take the mathematicians so long to prove such a simple and trivial statement in light of the simple foregoing observations and loads, loads more besides?"

I take this opportunity to thank you for your comment on my general proof of Fermat's theorem.I got read the main points in the CMNSEM paper by an expert{prof. In Number Theory).Proof is OK and elementary.Thank you again and may I kindly request you not to response to rascals.

To Thao TranIt looks like you're manipulating with the same equation. Here's example for n=3 (presented in standard form)(X+Y)^3-3(X+Y)^2*X+3(X+Y)*X^2-Z^3=0(X+Y)^3-3(X+Y)^2*Y+3(X+Y)*Y^2-Z^3=0After subtraction we obtain3(X+Y)^2(X-Y)=3(X+Y)(X^2-Y^2)

How can you be sure that we have(X+Y)^3-3(X+Y)^2*X+3(X+Y)*X^2-Z^3=0(X+Y)^3-3(X+Y)^2*Y+3(X+Y)*Y^2-Z^3=0 ?We, of course , always have (X+Y)^3-3(X+Y)^2*X+3(X+Y)*X^2-Z^3= C(X+Y)^3-3(X+Y)^2*Y+3(X+Y)*Y^2-Z^3=C.

Now,I understand why ITHINKWITHLIVER did condemn the elementary proof given by me, I suppose.This is the reason So far, every amateur has failed. With such puny methods, the chances you have succeeded are virtually nil. Jeffery Shalit.I have been also treated as an amateur.I solved a big problem in physics long ago. But I am a mathematician. Whatever, it is you(ITHINKWITHLIVER) have no right to use such words as mother f......If my proof is wrong, I don't know,it is up to anyone to tell it.I was too late in understanding the reason for ITHINKWITHLIVER'S comment.R.A.D..P

Unfortunately, it seems that the paper has been removed from the journal's website. I'm guessing it used to be the paper in volume 2, number 2, pp. 317-338. If anyone has a copy of the paper, please confirm the details. Thanks.

Not unfortunately ,fortunately all papers of mine are removed.I am responsible for this.Man like you can not understand the following. We researchers, do research for the sake of researches. Not for the foolish riff-raff or their existance.

You (joelreyesnoche)might be happy if the proof given in cmnsem vol.2 No.3 (2011)march p.57-63 (for your information) is wrong. Therefore may I ask Prof.Jefrrey Shalit. I am very much thankful to you Professor if you say weather my proof is correct or wrong.This would be great help to to me as well. RADP(Piyadasa Ranawaka)

To all that who that like to condemn the proof of Fermat's last theorem given by me.It is your ignorance or some other bad qualityWhenever I wanted to solve a big problem I registered for Ph.D or D.Sc at a standard university.In case of Fermat's last theorem I did the same This is actually to save the time I will tell later you why almost all amateur has been failed. Piyadasa Ranawaka

My interest in the Fermat Conjecture, (FC,) began as an interest in the Pythagorean theorem. I wasn't looking for other integral solutions to the n greater than 2 problem. I was more interested in the fact that odd integral values of 'a' resulted in 'b' being equal to (c-1.) Then, I began looking at the FC, and his statement, in the margin, that he had a solution, but it was too big to fit in the margin. I realized that with the mathematical tools in existence at that time, it was probably algebraic or geometric in expression. The solution proposed in 1995 that exceeded 120 pages, and used mathematics far in excess of what was available to Fermat was probably much more complex that needed. I began by using simple substitution to eliminate variables. It was determined that b= [(a to the 2)-1]/2, and c=[(a to the 2)+1]/2. Using an integral 'n', the problem became, "Is there a positive integer 'n', such that the equation: [(a to the n)-1]/n is an integer?" Although (a to the n) / n would result in an integral solution, because 1 is subtracted from the numerator prior to the division, there are NO positive integers for 'a' and 'n' greater than 2 that would result in b being positively non-fractional. COMMENTS? A.Conti, Allen@TheWalnutGrove.com or OWInc@att.net

I have found The Relations of Barlow when I read Ribenboim's Book. "Playing" with it, I found another sistem of equations, and then, I used modular congruences and I found what is summarized below (we can work just with natural numbers to analize the fermat's integer soluctions,because if one or 2 of the numbers are negatives, them we can change the place in the equation in order to have just positive numbers. If al of them are negative, we can multiply the equation by -1):

I didn't make an English version*, because I am having dificulties in the mahts expression (loses its format when I paste here), but if you follow the steps below, then you will achiev the same results :

* I published the entire demonstration in a Blog I created (in this blog is better to read the maths equations) and in other forums in Brazil and others countries :

Now,I will tell you why many many people are failed in the proof of Fermat's last theorem. Look carefully, all of them starts from the original equation x^n+y^n=Z^n and then look for a contradiction.This is utterly impossible since it has been assumed that x,y,z satisfy the equation. . I will give another example. In order to solve the elliptic curve y^2 +2=x^3, a famous equation , for integer roots one needs advanced methods such as unique factorization method if you starts from the original equation.Very good intelligence needs to do by elementary methods. One of my students has solved this equation using elementary mathematics.

To the best of my knowledge, there is no elementary proof of Fermat's last theorem for n=4 without depending on the Method of Infinite Descent of Fermat.I, with one of my students, designed an elementary proof without using any mathematical tool(only high-school Mathematics used).It was difficult and tricky.However, if we made a tiny mistake,then the proof followed at once.This is another evidence for my saying regarding the difficulty of proving Fermat's last theorem for all indices.Thank you very much Prof. Jefferey for maintaining this Blog, and wish you Merry Christmas in advance.

Thank you unknown. Would you offer such an award to the 17th century problems of finding all the integer integer roots of the simplified elliptic curves Y^2=x^3-2 , y^2+4=x^3 using which Fermat challenged the Europeans? I have already written the solutions using 17th century mathematics and Fermat's little theorem.

Before you go to a 'marginal proof' see weather you can prove Fermat's last theorem for n=4 without depending on Fermat's method of infinite descent.One of my students and myself have already done this, third one of that type.Not only this, but I have designed the proof of the FLT also.This is my third proof of Fermat's last theorem and this would be the proof of Fermat, who is actually ingenious.

I would not be worried about hatefull comments on somebody's short proofof Fermat's last theorem. I developed a short proof of integrability ofa continuous function defined on a closed finite interval. Proofs available in textbooks are usually 15 pages long in some cases a chapter. By the lengthof a proof I mean all the results one needs to prove the theorem starting from the definition of continuity. It is an interesting experience to see those idiotic reports from referees and letters from editors. I claim thatif somebody writes an idiotic report, he is an idiot. The question is if they are so stupid they do not understand the trouble of proving integrability. The answer is clear, they do not want shorter proofs. Let me ask them directly: "Why do you prefer long proofs to short ones?" This is a retoric question.I believe the same kind of people does not want to accept any short proofof FLT.Reference: Mathematics Student, last year.Sincerely,Josef Bukac

Fermat's last theorem have been made difficult by mathematicians.See how simple is the proof.Proof of the theoremLet us prove the theorem for n=4 first.proof of Fermat’s last theorem for n=4Fermat’s last theorem for n=4 can be stated thus;There are no non-trivial x,y,z integers satisfying the equation z^4=y^4+x^4,(x,y)=1 (1) Without loss of generality, we can assume that x,y,z>0 ,and also we have labeled the integer triple such that z>y>x>0 . Consider the equation z^4/x^4 =y^4/x^4 +1 which can be written as g^4=h^4+1, h= y/x>1 (2) where g,h are rational numbers .From (2),(g^2-h^2 )(g^2+h^2 )=1 (3) and let (g^2-h^2 )=d. Then g^2+h^2 =d+〖2h〗^2=1/d>0 . In other words,d>0 andd^2+2h^2 d-1=0 (4) We must have from (4) that 0<2h^2 d<1 (5) since〖 d〗^2>0,and d^2+2h^2 d=1〖00 This means that the quadratic in h^2 is positive and hence the discriminant ∆=〖4g〗^4-8<0 ,g^4<2 ,and therefore from (2), h^4<1.This is impossible since h= y/x>1. Hence, (1) has no non-trivial integer triple solution. After proving FLT for n=4 , one has to prove for FLT for any odd prime p in order to prove FLT in general[2]. Now, consider the equationz^p=y^p+x^p,(x,y)=1 (6) where p is any odd prime. By rearranging and relabeling of the components in (6), we can assume without loss generality that y>x>O. From(6), we obtain z^p/x^p =y^p/x^p +1 and this can be written as g^p=h^p+1, where g,h are positive and h>1 If y≠0. Definitions of g,h are obvious. Therefore g^p-h^p=1 and this gives(g-h)(g^(p-1)+〖hg〗^(p-2)+⋯…..+h^k g^(p-k-1)+⋯..+h^(p-2) g+h^(p-1) )=1 (7)Let g-h=d which is positive since the longer expression within bracket in (5) is positive, that is 1/d is positive. Then from (7),〖(h+d)〗^(p-1)+h〖(h+d)〗^(p-2)+……….+h^(p-2) (h+d)+h^(p-1)=1/d and this givesd^p+hd^(p-1)+⋯………+h^(p-2) d^2+ph^(p-1) d-1=0 (8) Note that all terms in (8) are positive except the constant term (-1). Now, we have h^(p-2) d^2+ph^(p-1) d-1<0 . In other words,〖0<(-h〗^(p-2))d^2-ph^(p-1) d+1 (9) and (9) is a quadratic in d if h≠0 and its discriminant should be negative. In other words∆<0,〖∆=p〗^2 h^(2(p-1))+4h^(p-2)<0 (10) This is impossible since h is positive, and therefore we conclude that there are no non-trivial x,y,z>0 integers satisfying the equation (6). Even if we consider the quadratic expression (9)as a function of g we will arrive at the same conclusion.May I ask the world not to make that Fermat's last theorem difficult by saying this and that.

Hello Unknown,I do not want to sell my research work for money whatever the amount may be.I know the proof is correct and i am addressing only the researchers of high potential to ask not to think that Fermat's last theorem is difficult and not to waste time by following the conventional approaches.

Thank you Prof.Jeffery.In case of n=4,I assumed there are non trivial integer triples and thereby new proof has been proposed.It is true that I have made a mistake in case of the index p.The equation last mentioned should be corrected as 〖d^p+⋯….+(p(p-1)/2)h〗^(p-2) d^2+p〖dh〗^(p-1))=1 However, arguments are the same.This is only a proof designed to show that proof of FLT is simple.

Hello Piyadasa, if you truly believe your proof to be correct, you should send it to a peer-reviewed journal. There are at least 20 to choose from in the area of number theory. Or if you just want feedback, for free, you can post it to the Yahoo! group associated with the Unsolved Problems site.

Dear Prof.Jeffery pl. let me add the corrected proof of FLT.Proof of the theoremproof of Fermat’s last theorem for n=4Fermat’s last theorem for n=4 can be stated thus;There are no non-trivial x,y,z integers satisfying the equation z^4=y^4+x^4,(x,y)=1 (1)Let us assume, on the contrary ,that there are x,y,z integers satisfying the above equation. Without loss of generality , we can assume that z>y>x>0.Now,consider the equation z^4/x^4 =y^4/x^4 +1,which can be written asg^4=h^4+1, h=y/x>1 (2)where g,h are rational numbers .From (2),(g^2-h^2 )(g^2+h^2 )=1 (3)and let (g^2-h^2 )=d. Then g^2+h^2=d+〖2h〗^2=1/d>0 . In other words,d>0 andd^2+2h^2 d-1=0 (4)We must have from(4) that0<2h^2 d<1 (5)since〖 d〗^2>0,andd^2+2h^2 d=1.〖00 This means that the quadratic in h^2 is positive and hence the discriminant ∆=〖4g〗^4-8<0,g^4<2,and therefore from (2),h^4<1.This is impossible since h=y/x>1. Hence, (1) has no non-trivial integer triple solution.Now, consider the equationz^p=y^p+x^p,(x,y)=1 (6)where p is any odd prime. As in the previous case, let us assume that there are x,y,z integers satisfying (6).Rearranging relabeling(6), we can make z>y>x>0.From (6), we obtain z^p/x^p =y^p/x^p +1 and this can be written as g^p=h^p+1, where g,h are positive and h>1.g^p-h^p=1 which gives(g-h)(g^(p-1)+〖hg〗^(p-2)+⋯…..+h^k g^(p-k-1)+⋯..+h^(p-2) g+h^(p-1) )=1 (7)Let g-h=d which is positive since the longerexpression within bracket in (7) is positive, that is,1/dis positive. Then from (7),〖(h+d)〗^(p-1)+h〖(h+d)〗^(p-2)+……….+h^(p-2) (h+d)+h^(p-1)=1/d and this givesd^p+hd^(p-1)+⋯………+h^(p-2) (p(p-1)/2)d^2+ph^(p-1) d-1=0 (8)Note that all terms in (8) except the constant term (-1) are positive. Since p is odd,(8) has one real root. Now, we have h^(p-2) 〖p(p-1)/2)d〗^2+ph^(p-1) d-1<0 .In other words,〖0<(-h〗^(p-2))(p(p-1)/2)d^2-ph^(p-1) d+1 which is a quadratic in d if h≠0 a its discriminant ∆〖∆=p〗^2 h^(2(p-1))+4.(p(p-1)/2)h^(p-2)<0 (9)This is impossible since h is positive, and therefore we conclude that there are no non-trivial x,y,z>0integers satisfying the equation (6). Even if we consider the quadratic expression is a function ofg we will arrive at the same conclusion.Thank you very much for your comment again.

Your reasoning for an odd prime is correct up until and including equation (8). Unfortunately, it then goes off the rails. The discriminant for a quadratic is b^2-4ac, not b^2+4ac. Both a and c are negative in this case. So your equation (9) is incorrect.

Unknown, don't be hurry. Note that all terms in (8) are positive except the constant term (-1). Now, we have h^(p-2) p(p-1)/2d^2+ph^(p-1) d-1<0 . In other words,〖0<(-h〗^(p-2))(p(p-1)/2)d^2-ph^(p-1) d+1 (9) and (9) is a quadratic in d if h≠0 and its discriminant should be negative. In other words∆<0,〖∆=p〗^2 h^(2(p-1))+4h^(p-2) {p(p-1)2}<0 (10)

I have obtained result b^2-4ac<0 from(9) Any way I am not surprised by your comment.

It is now clear that high school mathematics is sufficient to prove Fermat's last theorem.The way the proof has been done is completely different from the conventional approaches.the conventional approach: Assume that z^n=y^(n )+x^n,(x,y)=1,n>2 and try to draw a contradiction.This method was not successful for few hundred years.Fermat himself had a proof(I believe) similar to the proof we have suggested or a proof in which it is proved that one of x,y is zero after assuming the above equation.The saying of Andrew Wiles, I do not believe that he had a proof.He fooled himself to think he had a proof. has no foundation.

Challenge to the worldProof of Fermat's last theorem is very simple as i pointed out. Following the same argument to a certain extend and the very fundamental of algebraic number theory, I have already designed the third proof of Fermat's last theorem which is also very short.If any one doubt or challenge the proofs(at least given on the above) I am very happy. My job on Fermat's last theorem is not to gain anything but for the sake of mathematics is over.I thank prof. Jeffery for keeping this blog.

Simplest proof of Fermat's last theoremProof of the theorem Fermat’s last theorem for n(>2) can be stated thus: There are non-trivial integers x,y,z satisfying the equationz^n=y^n+x^n,(x,y)=1,n>2 (1) Rearranging and relabeling the integers, we can assume without loss of generality that z>y>x>0. From (1), we obtain g^n=h^n+1 (2) where g=z/x h= y/x. From the equation(2), we obtain〖(g-h)[g^(n-1)+g^(n-2) h+⋯……..+gh^(n-2)+h^(n-1)]=g〗^n-h^n=1, and if g-h=d , we have g^(n-1)+g^(n-2) h+⋯……..+gh^(n-2)+h^(n-1)=1/d>0 and therefore d>0.d([(d+h)^(n-1)+h(d+h)^(n-2)+⋯…….+(d+h) h^(n-2)+h^(n-1)])=1 Hence,d^n+nhd^(n-1)+⋯……+(n(n-1)/2) h^(n-2) d^2+nh^(n-1) d-1=0 (3) Since h,d>0,(n(n-1)/2) h^(n-2) d^2+nh^(n-1) d-1<0 in other words -(n(n-1)/2) h^(n-2) d^2-nh^(n-1) d+1>0 (4) (4) is a quadratic expression in d and is positive and therefore the discrininant should be negative ,i.e n^2 h^(2n-2)+2n(n-1) h^(n-2)<0 (5) which never holds and we conclude that (1) is not satisfied by non trivial integer triples x,y,z.

Probably, you are a troll, and realize that your "proof"is incorrect. However, I will give you the benefit of the doubt. In which case, does it not concern you that your "proof" never uses the fact that the quantities involved must be integers? So you have "proved" that there are no solutions to FLT even if the numbers involved are irrational. Which is, of course, nonsensical.

Hello,Unknown.What you say is complete nonsense. In the beginning, I have assumed that there are x,y,z>0 integers satisfying the equation X^n+y^n=z^n where n>2 and (x,y)=1 (1)After that I get the equation 1+h^n= g^n where h>1 (2) according to our fare assumption that 00 ,h cannot be never negative and this is a contradiction and it is sufficient to conclude that there are no positive integer solution for x^n+y^n=z^n . I have already proved that if there are positive integer solution for the Fermat equation (1) and only real solution we can have is h=0 and g=1.This means no positive integral y. Pl. understand that h=0 implies y=0.I do not argue with you hereafter. Thanks professor Geffry I am sorry for being late in answering.

I will be very happy if Andrew Wiles publishor republish his work in such a way that a High School Student can understand the proof.

You're going to be waiting a long time.

6:40 AM, December 31, 2013Thank you very much for the above comment.I have already proved and published in an international conference the proof which can be understood by an advanced level(AL) student in my country or high School student you have meant.Thank you so much. Long live CNMSEM? and that proof, you will find later.

I have designed a proof of Fermat's last theorem using the mathematics of Fermat's time. Read on the internet ' The Simplest proof of Fermat's last theorem,University of Kelaniya'Thank you all.Piyadasa Ranawaka

Fermat's last theorem can be proved using 17th century mathematics.What i have said is O.K.That is Fermat is ingenious is O.K I have been able to prove the theorem (Mean's theorem) g^p=h^p+1,where g=z/x ,h=y/x has only one rational solution only when g=1,h=0.That is corresponding Fermat equation has no non trivial integer triples. I think this is enough.Thank you prof. regarding your comments on the open access journals. However,Fermat's Last theorem,Beal's conjecture,Riemann Hypothesis have all been proved,in open access journals.I have read four papers.Wonderful!!.